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Calcium hydride reacts with water to form calcium hydroxide (aqueous) and hydrogen gas. How would you write a balanced chemical equation for the reaction? How many grams of calcium hydride are needed to form 4.600 of hydrogen?

Answer 1

Nolan Allen

#CaH_2(s) + 2H_2O(l) rarr Ca(OH)_2(aq) + H_2(g)uarr#

Approx. #100*g# calcium hydride are required.

If #4.6*g# of dihydrogen are evolved, this is a molar quantity of #(4.6*g)/(2.016*g*mol^-1)# #=# #2.28# #mol#.

If #2.28*mol# dihydrogen are evolved, then at least a an equivalent molar quantity of calcium hydride was used.

i.e. #2.28*molxx42.094*g*mol^-1" calcium hydride"# #=# #??g#

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Answer 2

Isabella Acosta

Sure, here are the answers:

Balanced chemical equation: [ \ce{CaH2 + 2H2O -> Ca(OH)2 + 2H2} ]
To find the grams of calcium hydride needed to form 4.600 g of hydrogen: [ \text{Molar mass of } \ce{CaH2} = 42.09 \text{ g/mol} ] [ \text{Molar mass of } \ce{H2} = 2.02 \text{ g/mol} ] [ \text{Moles of } \ce{H2} = \frac{4.600 \text{ g}}{2.02 \text{ g/mol}} \approx 2.277 \text{ mol} ] According to the balanced equation, 1 mol of (\ce{CaH2}) produces 2 mol of (\ce{H2}). So, moles of (\ce{CaH2}) needed = (2.277 \text{ mol} \times \frac{1 \text{ mol } \ce{CaH2}}{2 \text{ mol } \ce{H2}} = 1.138 \text{ mol}) [ \text{Mass of } \ce{CaH2} = 1.138 \text{ mol} \times 42.09 \text{ g/mol} \approx 47.93 \text{ g} ] So, approximately 47.93 grams of (\ce{CaH2}) are needed to form 4.600 grams of (\ce{H2}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.